# Relating equations to functions

1. Mar 30, 2013

### leehufford

1. The problem statement, all variables and given/known data

(1)I was hoping someone could help me understand this, this is for a Calc based Physics I class. I will post this problem, but its really the whole concept I'm not understanding. If someone can help me understand this concept, they need not solve the actual problem for me.

Given
θ(t) = a + bt - c(t^3)

a,b,c are constants

When t = 0, θ=pi/4 and ω = 2.00 rad/s
When t = 1.5s, α = 1.25 rad/s^2

Find a,b,c including their units.

2. Relevant equations

?? Angular speed is the 1st time derivative of θ?
?? Angular acceleration is the 2nd time derivative of θ?

3. The attempt at a solution

a and b are just the initial θ and ω. The t^3 is throwing me off. How does this all relate? Again, I don't want this problem done for me, I really want to get whats going on between equations (in this case rotation of rigid bodies) and mathematical polynomials.

One more thing... If you can answer the above you can probably answer this:
(2) Why does the angular velocity/acceleration vectors point into 3 dimensional space (z axis)?!?

Thank you so much for reading this!

-Lee

2. Mar 30, 2013

### Staff: Mentor

That makes three equations with three unknowns. You already solved two of them.

They are related to the axis of rotation. If rotational motion happens in a plane, then the axis of rotation is necessarily perpendicular to that plane.

3. Mar 30, 2013

### MalachiK

It seems like you've got a pretty good idea of what's going on here anyway. If you've worked out the values of a and b then you've differentiated to get the angular speed. Differentiate again and you'll get the acceleration. Sub in the values from the problem and you'll get c.

There may very well be something that I'm missing, but it seems like you've been given an arbitrary polynomial and just been tasked to differentiate it. The only significance of having a 3rd order equation is that you get a time dependence in both speed and acceleration. Anything less than that would give you an acceleration that doesn't change with time. I suppose that this is to make the problem a bit more interesting.

As for the direction of the angular acceleration vectors - the best I can do for you is to say that you can get ω from a vector cross product of the tangential velocity and the radius and the cross product is defined so that the answer is always perpendicular to both of the vectors that you started with. I'm sure someone else might be able to give you a more satisfying answer.

4. Mar 30, 2013

### leehufford

θ(t) = a + bt - c(t^3)
ω(t) = b - 3c(t^2)
α(t) = -6ct

Given that at t=1.5s, α=1.25 rad/s^2 I substitute:

Which is the correct value with the correct units. How can angular acceleration have units of rad/s^3? I still don't undertstand how this actually works. Thanks for the reply and all future replies.

-Lee

5. Mar 30, 2013

### leehufford

Thank you for the reply. We were just introduced to the cross product a week or so ago when talking about Torque. I sort of understand how that works now. My big problem really is how these polynomials relate to physics equations.. because a polynomial is for only x, yet these polynomials seem to have θ, ω, and α in them. I really don't even see why a and b are what they are.

Last edited: Mar 30, 2013
6. Mar 30, 2013

### Staff: Mentor

You wrote it yourself: the angular acceleration (α) has units of rad/s^2.

7. Mar 30, 2013

### cepheid

Staff Emeritus
Yeah, this is correct. Why the uncertainty?

I will try to help, but this is a REALLY vague question. Could you try asking a specific one?

The angular displacement θ is a cubic function of time, which means that if you plot θ vs. t, starting at t = 0 you get something like what's shown in the attached plot. That's all. What is it about the functional form being a cubic that is throwing you off?

It's just a definition. Quantities like angular momentum and torque are treated like vectors, with the relationship between the direction of the vector and the direction of the associated rotation being given by the right hand rule. This is useful because these quantities have many of the mathematical properties of vectors. They can add like vectors etc. However, they do not share all of the properties of vectors: they are pseudo-vectors.

You might have seen the demo (or tried it yourself) where you have a vertical bicycle wheel that is spinning (oriented such that you are viewing it edge-on, and its axis goes from left to right, from your point of view). You are holding it from that axis (or axle), one hand on one side, and one hand on the other side. You are sitting in a swivel chair. You try to rotate the bicycle wheel away from vertical around an axis perpendicular to the axle, so that it goes from being oriented like this: ||, to being oriented like this: =

Much to your surprise, when you try to rotate the wheel in this fashion, the whole system (you + wheel) ends up rotating around the axis of the swivel chair (which is perpendicular to the other two axes)! At first this seems pretty wacky and hard to understand, but if you just consider the directions of the angular momentum and torque, and sum the vectors, it all makes sense. That's why the vector formalism is useful for this type of quantity. Makes is easy to try to add rotations around different axes together (which is otherwise hard to visualize). Look up gyroscopic precession for more details.

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8. Mar 30, 2013

### Staff: Mentor

You have an equation for the angle as a function of time. This equation is posited to be valid for any time t. Therefore, the equation must somehow contain the initial angle, angular speed and angular acceleration.

9. Mar 30, 2013

### cepheid

Staff Emeritus
Oh please, polynomials are not just "for x." The letter x is just a symbol denoting a VARIABLE. You could have used any symbol, like q or w or δ or . In physics we deal with variables. Functions are just relationships between two variables, a dependent one, and and independent one. Here, the independent variable is TIME (t), and the dependent variable is the ANGLE (θ). So, θ DEPENDS on t. It is a function of t. In this case, it happens to be a polynomial function of t. So we have t instead of x. Is that really what's throwing you off, that the "x" in this polynomial has a physical meaning, that it is some measured quantity? Because that's true of all mathematical relations in physics. The quantities a, b, and c are the COEFFICIENTS of the polynomial, just like usual.

10. Mar 30, 2013

### leehufford

Thanks for your reply. I think what was tripping me up is that I thought angular acceleration was constant, but clearly it isn't becuase the function for angular acceleration still has a variable in it. I see now that acceleration is not constant. Also, I just noticed that the units of each term must simplify to theta, so the units of rad/s^3 makes sense.

If any of my assumptions here are wrong please let me know before I think I'm right!! Thanks again for all replies.

-Lee

11. Mar 30, 2013

### leehufford

See I thought a,b,c where more than coefficients. I thought I was dealing with a polynomial of three different variables for a minute. It is still weird that coefficients have units attached to them though. Thanks for the reply.

-Lee

12. Mar 30, 2013

### cepheid

Staff Emeritus
Nope, as you have discovered, a, b, and c are coefficients, that are determined by the initial conditions of the system. The coeff. 'a' is the initial position, 'b' is the initial angular speed, and 'c' is proportional to the rate of change (derivative) of alpha. That's why they have units. Basically, in this problem, alpha is not constant, but varies linearly with time, with a slope -6c.

It may be that many of the examples you considered has a constant angular acceleration (alpha) (In which case θ vs. t would be quadratic). But that does not mean that alpha HAS to be constant. There is no reason why the acceleration can't vary with time. (In fact, it has to, if the net torque varies with time).

Regarding rad/s3, if you are talking about the units for 'c', then I agree. Your reasoning is correct about every term having to have units of rad. Another way to look at it: as I said above, c is the rate of change of alpha, which means it would have dimensions of acceleration/time.

13. Mar 30, 2013

### leehufford

It makes sense now. I was talking about c. I've learned a lot. Thanks!